Download HOMEWORK 2: SOLUTIONS - MATH 110 INSTRUCTOR: George

HOMEWORK 2: SOLUTIONS - MATH 110
INSTRUCTOR: George Voutsadakis
Problem 1 Suppose that you were able to take a large piece of paper of ordinary thickness
and fold it in half 50 times. What is the height of the folded paper? (Look on the web for a
reasonable estimate of the ”ordinary” thickness.)
Solution:
Suppose that you found that the thickness of the paper is a inches. If you fold the paper
1 time you will have 2 sheets on top of each other. If you repeat, i.e., fold it 2 times, you
will have 4 sheets on top of each other. For 3 folds, you will have 8 sheets on top of each
other. For 4 folds, you get a 16 sheets thick bunch. The pattern now should be becoming
clear. If we fold the paper n times, we’ll have a 2n sheets thick bunch. Thus the number
we were asked to compute should be
a × 250 .
Problem 2 Suppose that 370 people are attending a party? Are there two people in the
party having the same birthday?
Solution:
Think of the different possible birthdays as boxes: 1/1, 1/2, 1/3, . . . , 2/1, . . . , 12/31.
Then place each of the 370 people in the box that is labelled by his/her birthday. By
the pigeon-hole principle, at least one box will have to contain at least two people. Thus,
there will surely be at least two people with the same birthday at the party.
Problem 3 What proportion of the first 1,000 numbers have a 3 somewhere in them?
What proportion of the first 10,000 numbers have a 3? Explain why almost all million-digit
numbers contain a 3.
Solution:
It is much easier to compute the number of numbers that do not contain a 3. In the
first 10 numbers 9 do not contain a 3. In the first 100 numbers 81 do not contain a 3. (This
is because these numbers are formed by selecting a digit out of {0, 1, 2, 4, 5, 6, 7, 8, 9} for
the first position and another digit out of the same set for the last position and this may
be done in 9 × 9 = 81 ways. In the first 1000 numbers, reasoning as above, there are 9 3
numbers that do not contain a 3. Now the pattern should be becoming obvious. In the first
10n numbers there are 9n that do not contain a 3. This shows that the number of those
numbers in the first 10n that do contain a 3 is 10n − 9n . Thus, the ratio of those numbers
that do contain a 3 over the total is
10n − 9n
10n
9n
9
=
−
= 1 − ( )n .
n
n
n
10
10
10
10
Verify that this goes to 1 as n gets very large. Therefore almost all million digit numbers
do contain a 3.
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Problem 4 By experimenting with numerous examples in search of a pattern, determine
a simple formula for (Fn+1 )2 + (Fn )2 , that is a formula for the sum of the squares of two
consecutive Fibonacci numbers.
Solution:
We have
n
Fn
Fn2
Fn2
2
Fn+1
2
+ Fn+1
0
1
1
1
2
1 2 3 4
5
6
7
8
9
10
1 2 3 5
8 13
21
34
55
89
1 4 9 25 64 169 441 1156 3025 7921
4 9 25 64 169 441 1156 3025 7921
5 13 34 89 233 610 1597 4181 10946
From the table above, it is clear that F02 + F12 = F2 , F12 + F22 = F4 , F22 + F32 = F6 , F32 + F42 =
F82 , . . . . Thus, the pattern that emerges is
2
Fn2 + Fn+1
= F2(n+1) .
Problem 5 By experimenting with numerous examples in search of a pattern, determine
a formula for Fn + Ln , that is a formula for the sum of a Fibonacci number and the corresponding Lucas number.
Solution:
We have
n
Fn
Ln
Fn + L n
2Fn + Fn−2
0
1
2
3
1
1
1
2
2
2
3
5
5
3 4 5
3 5 8
4 7 11
7 12 19
7 12 19
6
13
18
31
31
7
21
29
50
50
8
9 10
34 55 89
47 76 123
81 131 212
81 131 212
From the table above, it is clear that the pattern that emerges is
Fn + Ln = 2Fn + Fn−2 ,
for n ≥ 2.
Problem 6 Let’s start with the numbers 0,0,1 and generate future numbers in our sequence
by adding up the the previous three numbers. Write out the first 15 terms in this sequence,
starting with the first 1. Evaluate the values of the quotients of consecutive terms (dividing
the smaller term into the larger one). Do the quotients seem to be approaching a fixed
number?
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Solution:
We have
n
Sn
Sn+1
Sn
Sn+1
n
Sn
Sn+1
Sn
Sn+1
0 1 2
3
4
5
6
7
8
0 0 1
1
2
4
7
13
24
0 1 1
2
4
7
13
24
1 0.5 0.5 .571428 .538461 .541666 .545454
9
10
11
12
13
14
15
16
17
44
81
149
274
504
927
1705
3136
44
81
149
274
504
927
1705
3136 5768
.543209 .543624 .543795 .543650 .543689 .543695 .543686 543689
We observe that the numbers are indeed approaching a specific number.
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